Question

divide.
$(4x^3 - 4x^2 - 6x - 9) \div (2x - 4)$
your answer should give the quotient and the remainder.
quotient:
remainder:

Explanation:

Step1: Use polynomial long division

Divide the leading term of the dividend \(4x^3\) by the leading term of the divisor \(2x\), we get \(2x^2\). Multiply the divisor \((2x - 4)\) by \(2x^2\) to get \(4x^3-8x^2\). Subtract this from the dividend: \((4x^3 - 4x^2 - 6x - 9)-(4x^3 - 8x^2)=4x^2 - 6x - 9\).

Step2: Divide the new leading term

Divide the leading term of the new dividend \(4x^2\) by \(2x\), we get \(2x\). Multiply the divisor \((2x - 4)\) by \(2x\) to get \(4x^2-8x\). Subtract this from the new dividend: \((4x^2 - 6x - 9)-(4x^2 - 8x)=2x - 9\).

Step3: Divide the new leading term again

Divide the leading term of the new dividend \(2x\) by \(2x\), we get \(1\). Multiply the divisor \((2x - 4)\) by \(1\) to get \(2x - 4\). Subtract this from the new dividend: \((2x - 9)-(2x - 4)=-5\).

Answer:

Quotient: \(2x^2 + 2x + 1\)
Remainder: \(-5\)